[r6rs-discuss] eqv? seems unwell

From: Alan Watson <alan>
Date: Wed, 27 Jun 2007 18:16:54 -0500

Alan Watson wrote:
> I know of at least two R5RS
> implementations in which the following need not evaluate to #t:

It gets worse.

I also know of lots and lots of R5RS implementations in which the
following WILL NOT evaluate to #t:

   (let* ((x 1.0)
          (foo (lambda (y) (eq? x y)))
          (obj1 x)
          (obj2 (+ 0.0 x)))
     (eqv? (foo obj1) (foo obj2)))

Normally, all that's needed here is that the implementation use boxed
inexacts and doesn't share them on the fly.

Regards,

Alan
Received on Wed Jun 27 2007 - 19:16:54 UTC